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I have been attempting to get my mind around the Kuhn-Munkres/Hungarian Algorithm. I have been using the following statement of the algorithm which I found here.

Kuhn-Munkres Algorithm

Based on my readings on the algorithm, my understanding is that the improvement of the feasible vertex labeling in step 2. is supposed to be such that $G_l \subset G_l'$.

The part I'm getting stuck on is that it seems to me that it is possible for there to be an edge $xy$ in $G_l$ with $y \in T$ but $x \not \in S$ resulting in the fact that $$ l'(x) + l'(y) = l(x) + (l(y) + \alpha_l) \not = w(xy) $$ and so $ xy \not \in G_l'$ and $G_l \not \subset G_l'$.

Can anyone point out where I'm going wrong?

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    $\begingroup$ Please copy the text from the picture here, so it can be searched for $\endgroup$ – Raphael♦ Dec 12 '12 at 20:25
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    $\begingroup$ Mere transcribing wouldn't do the trick, it would need quite a bit of mathjaxing which I'm not really up for. But that's beside the point, any potentially searchable 'buzzwords' appear in the body of the question as text. $\endgroup$ – Isaac Kleinman Dec 13 '12 at 15:39
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    $\begingroup$ I'm afraid "I don't want to" is not an option here. Posts whose content is buried in images are generally bad form, to the point of being deletion-worthy (see here, here). Therefore, you should really invest the work (what's the message if you don't?). See here for a primer on how to use maths here. $\endgroup$ – Raphael♦ Dec 13 '12 at 20:02
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    $\begingroup$ you fail to address my main point that anything anyone would search for is NOT embedded in the image, so unless someone does a search for mathematical jargon, I think they'll be fine. $\endgroup$ – Isaac Kleinman Dec 14 '12 at 15:32
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    $\begingroup$ I think this discussion is going on for longer than it should, but take a look at the image: subtract all words that appear as text in my question; also, subtract all mathematical markup since that's not searchable anyway. Now, what words are left, 'choose', 'find', 'otherwise'. Do we really have to accommodate someone searching for those terms? $\endgroup$ – Isaac Kleinman Dec 19 '12 at 18:14
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You are right that $G_{l'}$ is not necessarily a superset of $G_l$. However, you can still prove that the algorithm runs in strongly polynomial time. You have the following invariants:

  • all edges in $M$ remain edges of $G_{l'}$
  • we don't remove vertices from $S$ and $T$ until we increase the size of the matching $M$, at which point $S$ and $T$ are reset (step 1); so the size of $S$ and $T$ is monotonically increasing until the size of $M$ is increased by 1
  • after updating the labels to $l'$, the size of $T$ is increased by at least 1 in the next step

What can you conclude from this? The size of the matching $M$ never decreases. At each iteration we either increase the size of $T$, or we update the labels, which will cause us to increase the size of $T$ in the next iteration. So after $2n$ iterations, the size of $T$ will be $n$. Since $T$ cannot grow anymore, we will have to increase the size of $M$. But the size of $M$ is at most $n$, so the algorithm will finish after at most $O(n^2)$ iterations. An iteration can be executed in time $O(m)$, so the total running time is bounded by $O(n^2m)$.

BTW the sets $S$ and $T$ are a bit mysterious in this description of the algorithm. Here is how we usually think about them. Orient all edges in $G_l$ as follows: the edges in $M$ go from $Y$ to $X$ and all other edges go from $X$ to $Y$. Then $x$ is an unmatched vertex, $T$ is computed to be the set of vertices in $Y$ reachable from $x$ by a directed path, and $S$ is the set of vertices in $X$ reachable from $x$ by a directed path. If $T$ contains an unmatched vertex, we have found an odd-length alternating path, and we can augment the matching (increase its size by reversing the direction of the edges along the path). Otherwise, we can change $l$ so that the size of $T$ increases.

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